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Question
The ratio between the de Broglie wavelength associated with proton accelerated through a potential of 512 V and that of alpha particle accelerated through a potential of X volts is found to be one. Find the value of X.
Solution
de-Broglie wavelength of accelerated charge particle
λ = `"h"/sqrt(2"mqV")`
`λ ∝ "h"/sqrt("mqV")`
Ratio of wavelength of proton and alpha particle
`λ_"p"/λ_α = sqrt(("m"_α"q"_α"V"_α)/("m"_"p""q"_"p""V"_"p")) = sqrt(("m"_α/"m"_"p") ("q"_α/"q"_"p") ("V"_α/"V"_"p"))`
Here, `"m"_α/"m"_"p"` = 4; `"q"_α/"q"_"p"` = 2; `"V"_α/"V"_"p" = "X"/512`; `λ_"p"/λ_α` = 1
1 = `sqrt(4 xx 2 xx ("X"/512))`
= `sqrt("X"/64)`
= `"X"/64`
X = 64 V
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